define定义函数

2025-01-22 22:53:56
推荐回答(4个)
回答1:

#define Abs(x) (x>=0 ? x:(-x))
宏定义只是完全替换,不会自动给你加括号的
Abs(a-b)变成
Abs(a-b) (a-b>=0 ? a-b:(-a-b))
-a-b 就出现问题了,负号的优先级比减号高的

所以在宏定义里,每项运算都要加括号,因为括号的优先级最高,保证表达式的运算顺序不会发生改变
#define Abs(x) ((x)>=0 ? (x):(-(x)))

回答2:

这是带参数的宏,不是函数,编译预处理的结果是字符替换,替换成如下形式:
(a-b>=0?a-b:(-a-b))
结果在a-b<0时并不正确,直接定义函数,当然如果是C,可能有数据类型的问题,如果是C++,可以定义一个模板
double Abs(double x)
{
return x >= 0 ? x : -x;
}

回答3:

对于类似a-b表达式的情况,两种解决办法:
1. 引用的时候用 Abs((a-b)) 就可以了,加一对括号,将表达式看成一个量,就不会出错了
2.改define,每项都加括号 #define Abs(x) ((x)>=0 ? (x):(-(x)))

回答4:

  c语言允许宏带有参数。在宏定义中的参数称为形式参数,在宏调用中的参数称为实际参数。对带参数的宏,在调用中,不仅要宏展开,而且要用实参去代换形参。
  带参宏定义的一般形式为:  #define 宏名(形参表) 字符串
  在字符串中含有各个形参。 带参宏调用的一般形式为:  宏名(形参表)
  例如:
  #defineM(y)((y)*(y)+3*(y))/*宏定义*/
  k=M(5);/*宏调用*/
  在宏调用时,用实参5去代替形参y,经预处理宏展开后的语句为:  k=5*5+3*5
  程序2:
  //#include
  #include
  #define MAX(a,b)((a>b)?(a):(b))
  intmain(intargc,char*argv[])
  {
  intx,y,max;
  printf("inputtwonumbers:");
  scanf("%d%d",&x,&y);
  max=MAX(x,y);
  printf("max=%d\n",max);
  return0;
  }
  上例程序的第一行进行带参宏定义,用宏名MAX表示条件表达式 (a>b)?a:b ,形参a,b均出现在条件表达式中。程序中 max=MAX(x,y) 为宏调用,实参x,y,将代换形参a,b。宏展开后该语句为: max=(x>y)?x:y;  用于计算x,y中的大数。
  #define 条件编译
  头文件(.h)可以被头文件或C文件包含;重复包含(重复定义)由于头文件包含可以嵌套,那么C文件就有可能包含多次同一个头文件,就可能出现重复定义的问题的。 通过条件编译开关来避免重复包含(重复定义)

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